Question

Ayuda con esas preguntas de química, si no tienen la respuesta de todas al menos con que tengan una
se los agradecería muchísimo

P.D la que más me interesa es las dos y cuatro

Answer 1

1.

Mm NaOH = 40 g/mol

       Ca(OH)2 = 74 g/mol

g NaOH= 500 g Ca(OH)2 x 1 mol Ca(OH)2 x 2 mol NaOH x 40 g NaOH

                                           ````````````````````      `````````````````` ````````````````

                                           74 g Ca(OH)2        1 mol Ca(OH)2  1 mol NaOH

g = 540.54 de NaOH

2. Mm  Zn3(PO4)2 = 386 g/mol
                         Zn = 65.38 g/mol

g Zn3(PO4)2 = 10 g Zn x  1 mol Zn  x  1 mol Zn3(PO4)2  x  386 g Zn3(PO4)2
                                         ``````````````    ````````````````````````   ```````````````````````
                                          65.38 g Zn        3 mol Zn              1 mol Zn3(PO4)2

g = 19.68 g de Zn3(PO4)2

3.  Mm C3H8 = 44 g/mol
                O2 = 32 g/mol

calcular moles a partir de los gramos

moles C3H8 = 20 g/44 g/mol = 0.454 mol
moles O2 = 20 g / 32 g/mol = 0.625 mol

calcular moles a partir de la reacción

n CO2 = 0.454 mol C3H8  x  3 mol CO2
                                              ````````````````
                                              1 mol C3H8

nCO2 = 1.362 mol

n CO2 = 0.625 mol O2  x    3 mol CO2
                                            ```````````````
                                              5 mol O2
n CO2 = 0.375

Reativo limitante: O2 produce 0.375 moles de CO2

Calcular reactivo en exceso:

g C3H8 = 20 g O2  x  1 mol O2  x  1 mol C3H8  x  44 g C3H8
                                  ``````````````     ````````````````      ```````````````
                                    32 g O2         5 mol O2         1 mol C3H8

g C3H8 = 5.5 g

g exceso = 20 g - 5.5 g = 14.5 g de C3H8

4. Mm AlBr3 = 267 g/mol
                Al = 27 g/mol
                Br2 = 160 g/mol

1. calcular moles a partir de los gramos

moles Al = 25 g / 27 g/mol = 0.926 mol Al
moles Br2 = 100 g / 160 g/mol = 0.625 mol

2. calcular moles de AlBr3

moles AlBr3 = 0.926 moles Al  x  2 mol AlBr3
                                                    ``````````````````
                                                        2 mol Al

moles AlBr3 = 0.926

moles AlBr3 = 0.625 moles Br2  x  2 mol AlBr3
                                                      ``````````````````
                                                         3 mol Br2

moles AlBr3 = 0.416 mol

REACTIVO LIMITANTE: Br2

calcular masa de AlBr3

m = 0.416 mol x 267 g/mol  = 111.07 g AlBr3

calcular % de rendimiento

% rend. = 64.2 g / 111.07 g  x  100

% rend. = 57.80 %